velocity-time graph of a block of mass 100 g sliding on a horizontal concrete force of 5 N is shown below. The magnitude of frictional force acting on the block due to the floor is
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5N since frictional force is opposing force
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Here is your solution
Given :-
U = 60m/s
V= 80m/s
Time = 8 sec- 6 sec = 2 sec
acceleration =v-u/t
a=>80-60/2=
a=>10m/s
mass 100 gm = 0.1kg
Force = mass ×acceleration
f=>0.1×10= 1N
constant force 5 newton( Given )
Net force = constant force - frictional force
N.f= 5 - frictional force
1 -5 = - frictional force
frictional force = 4 newton
Hope it helps you
Given :-
U = 60m/s
V= 80m/s
Time = 8 sec- 6 sec = 2 sec
acceleration =v-u/t
a=>80-60/2=
a=>10m/s
mass 100 gm = 0.1kg
Force = mass ×acceleration
f=>0.1×10= 1N
constant force 5 newton( Given )
Net force = constant force - frictional force
N.f= 5 - frictional force
1 -5 = - frictional force
frictional force = 4 newton
Hope it helps you
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